CODEWITHSUNDEEP
c++
Shreyas
Shreyas

Reputation: 36

How do I convert a string containing one number into a vector in C++ where all the digits are the elements of vector

Suppose I have a string containing 00101 and I need to convert it to a vector containing each single digit as an element, like vector<int> num{0,0,1,0,1}.

I looked up many solutions but the numbers in the string were space-separated.

Upvotes: 1

Views: 108

Answers (3)

janekb04
janekb04

Reputation: 4985

You can easily do this by initializing a new vector with an iterator pair like this:

std::string s = "2345676543456789"; // for example

std::vector<int> nums {s.begin(), s.end()};
for (auto&& num : nums)
    num -= '0';

If you now print the vector, like this:

for (const auto& n : nums) 
     std::cout << n << ' ';

it will contain all of the digits:

2 3 4 5 6 7 6 5 4 3 4 5 6 7 8 9

Working demo link.

Note that this breaks if the source string (s in this case) isn't composed only of digits. This is because this solution is based on the fact that in ASCII, digits are encoded with consecutive values.

Upvotes: 4

t.niese
t.niese

Reputation: 40882

You can use std::transform and and std::back_inserter.

c - '0' converts textual representation number to its value.

#include <algorithm>
#include <iostream>
#include <string>
#include <vector>

int main() {
  std::string s = "00101";

  std::vector<unsigned short> digits;
  std::transform(s.begin(), s.end(), std::back_inserter(digits),
                 [](const auto &c) { return c - '0'; });

  for (const auto &digit : digits) {
    std::cout << digit << std::endl;
  }

  return 0;
}

You could also get rid of back_inserter by initializing the std::vector with a set of elements.

#include <algorithm>
#include <iostream>
#include <string>
#include <vector>

int main() {
  std::string s = "00101";

  std::vector<unsigned short> digits(s.size());
  std::transform(s.begin(), s.end(), digits.begin(),
                 [](const auto &c) { return c - '0'; });

  for (const auto &digit : digits) {
    std::cout << digit << std::endl;
  }

  return 0;
}

Upvotes: 1

iyers16
iyers16

Reputation: 42

It may not be the best solution...but a solution nonetheless, please feel free to reply and improve my code:

#include <iostream>
#include <vector>
#include <string>
using namespace std;
int main()
{
    vector<int> v;
    string s = "101010";
    
    for(char c : s)
    {
        v.push_back(((int)c - 48));
    }
    
    for(int i = 0; i < v.size(); i++)
    {
        cout << v.at(i) << endl;
    }
    return 0;
}

The cout at the end isn't essential, just to see that the vector truly holds all the needed values.

As for the - 48 while pushing to the vector is because number chars in the ASCII table are denoted from the number 48 onwards, so to subtract the numerical values, we get the originally needed int. EDIT: Instead of subtracting from the typecast, we can also use the notation:

for(char c : s)
    {
        v.push_back(c - '0');
    }

which performs exactly the same thing but in a more readable way. This will bring a bit more clarity into why this is.

Upvotes: 1

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